simplifies to the characteristic equation . Because the equation factors into there is only one eigenvalue .

; ,

; ,

;

;

This exercise is recommended for all readers.

Problem 2

For each matrix, find the characteristic equation, and the eigenvalues and associated eigenvectors.

Answer

The characteristic equation is . Its roots, the eigenvalues, are and . For the eigenvectors we consider this equation.

For the eigenvector associated with , we consider the resulting linear system.

The eigenspace is the set of vectors whose second component is twice the first component.

(Here, the parameter is only because that is the variable that is free in the above system.) Hence, this is an eigenvector associated with the eigenvalue .

Finding an eigenvector associated with is similar. This system

leads to the set of vectors whose first component is zero.

And so this is an eigenvector associated with .

The characteristic equation is

and so the eigenvalues are and . To find eigenvectors, consider this system.

For we get

leading to this eigenspace and eigenvector.

For the system is

leading to this.

Problem 3

Find the characteristic equation, and the eigenvalues and associated eigenvectors for this matrix. Hint. The eigenvalues are complex.

Answer

The characteristic equation

has the complex roots and . This system

For Gauss' method gives this reduction.

(For the calculation in the lower right get a common denominator

to see that it gives a equation.) These are the resulting eigenspace and eigenvector.

For the system

leads to this.

Problem 4

Find the characteristic polynomial, the eigenvalues, and the associated eigenvectors of this matrix.

Answer

The characteristic equation is

and so the eigenvalues are (this is a repeated root of the equation) and . For the rest, consider this system.

When then the solution set is this eigenspace.

When then the solution set is this eigenspace.

So these are eigenvectors associated with and .

This exercise is recommended for all readers.

Problem 5

For each matrix, find the characteristic equation, and the eigenvalues and associated eigenvectors.

Answer

The characteristic equation is

and so the eigenvalues are and also the repeated eigenvalue . To find eigenvectors, consider this system.

For we get

leading to this eigenspace and eigenvector.

For the system is

leading to this.

The characteristic equation is

and the eigenvalues are and (by using the quadratic equation) and . To find eigenvectors, consider this system.

Substituting gives the system

leading to this eigenspace and eigenvector.

Substituting gives the system

(the middle coefficient in the third equation equals the number ; find a common denominator of and then rationalize the denominator by multiplying the top and bottom of the frsction by )

which leads to this eigenspace and eigenvector.

Finally, substituting gives the system

which gives this eigenspace and eigenvector.

This exercise is recommended for all readers.

Problem 6

Let be

Find its eigenvalues and the associated eigenvectors.

Answer

With respect to the natural basis the matrix representation is this.

Thus the characteristic equation

is . To find the associated eigenvectors, consider this system.

Plugging in gives

Problem 7

Find the eigenvalues and eigenvectors of this map .

Answer

, ,

This exercise is recommended for all readers.

Problem 8

Find the eigenvalues and associated eigenvectors of the differentiation operator .

Answer

Fix the natural basis . The map's action is , , , and and its representation is easy to compute.

We find the eigenvalues with this computation.

Thus the map has the single eigenvalue . To find the associated eigenvectors, we solve

to get this eigenspace.

Problem 9

Prove that

the eigenvalues of a triangular matrix (upper or lower triangular) are the entries on the diagonal.

Answer

The determinant of the triangular matrix is the product down the diagonal, and so it factors into the product of the terms .

This exercise is recommended for all readers.

Problem 10

Find the formula for the characteristic polynomial of a matrix.

Answer

Just expand the determinant of .

Problem 11

Prove that the characteristic polynomial of a transformation is well-defined.

Answer

Any two representations of that transformation are similar, and similar matrices have the same characteristic polynomial.

This exercise is recommended for all readers.

Problem 12

Can any non- vector in any nontrivial vector space be a eigenvector? That is, given a from a nontrivial , is there a transformation and a scalar such that ?

Given a scalar , can any non- vector in any nontrivial vector space be an eigenvector associated with the eigenvalue ?

Answer

Yes, use and the identity map.

Yes, use the transformation that multiplies by .

This exercise is recommended for all readers.

Problem 13

Suppose that and . Prove that the eigenvectors of associated with are the non- vectors in the kernel of the map represented (with respect to the same bases) by .

Answer

If then under the map .

Problem 14

Prove that if are all integers and then

has integral eigenvalues, namely and .

Answer

The characteristic equation

simplifies to . Checking that the values and satisfy the equation (under the condition) is routine.

This exercise is recommended for all readers.

Problem 15

Prove that if is nonsingular and has eigenvalues then has eigenvalues . Is the converse true?

Answer

Consider an eigenspace . Any is the image of some (namely, ). Thus, on (which is a nontrivial subspace) the action of is , and so is an eigenvalue of .

This exercise is recommended for all readers.

Problem 16

Suppose that is and are scalars.

Prove that if has the eigenvalue with an associated eigenvector then is an eigenvector of associated with eigenvalue .

Prove that if is diagonalizable then so is .

Answer

We have .

Suppose that is diagonal. Then is also diagonal.

This exercise is recommended for all readers.

Problem 17

Show that is an eigenvalue of if and only if the map represented by is not an isomorphism.

Answer

The scalar is an eigenvalue if and only if the transformation is singular. A transformation is singular if and only if it is not an isomorphism (that is, a transformation is an isomorphism if and only if it is nonsingular).

Problem 18

Show that if is an eigenvalue of then is an eigenvalue of .

What is wrong with this proof generalizing that? "If is an eigenvalue of and is an eigenvalue for , then is an eigenvalue for , for, if and then "?

Where the eigenvalue is associated with the eigenvector then . (The full details can be put in by doing induction on .)

The eigenvector associated wih might not be an eigenvector associated with .

Problem 19

Do matrix-equivalent matrices have the same eigenvalues?

Answer

No. These are two same-sized, equal rank, matrices with different eigenvalues.

Problem 20

Show that a square matrix with real entries and an odd number of rows has at least one real eigenvalue.

Answer

The characteristic polynomial has an odd power and so has at least one real root.

Problem 21

Diagonalize.

Answer

The characteristic polynomial has distinct roots , , and . Thus the matrix can be diagonalized into this form.

Problem 22

Suppose that is a nonsingular matrix. Show that the similarity transformation map sending is an isomorphism.

Answer

We must show that it is one-to-one and onto, and that it respects the operations of matrix addition and scalar multiplication.

To show that it is one-to-one, suppose that , that is, suppose that , and note that multiplying both sides on the left by and on the right by gives that . To show that it is onto, consider and observe that .

The map preserves matrix addition since follows from properties of matrix multiplication and addition that we have seen. Scalar multiplication is similar: .

? Problem 23

Show that if is an square matrix and each row (column) sums to then is a characteristic root of . (Morrison 1967)

Answer

This is how the answer was given in the cited source.

If the argument of the characteristic function of is set equal to , adding the first rows (columns) to the th row (column) yields a determinant whose th row (column) is zero. Thus is a characteristic root of .